是：a=(√3,1),=π/3,|b|=1是吧?
1
令b=(x,y),則：x^2+y^2=1
a·b=|a|*|b|*cos
=2*cos(π/3)=1
即：a·b=(√3,1)·(x,y)=√3x+y=1
即：(1-√3x)^2+x^2=4x^2-2√3x+1=1
即：x=0或√3/2
即：y=1或-1/2
故：b=(0,1)或(√3/2,-1/2)
2
c=(-√3,1),b、c共線
故：b=(√3/2,-1/2)
b⊥d,即：b·d=(√3/2,-1/2)·(√3x^2,x-y^2)
=3x^2/2-x/2+y^2/2=0
即：y^2=-3x^2+x
故：t=y^2+5x+4=-3x^2+x+5x+4
=-3x^2+6x+4
=-3(x-1)^2+7
當(dāng)x=1時(shí),y取得最大值：7